Question

help

Answer 1

...

Answer 2

I'm having trouble with this one

Answer 3

don't be confused
just go with what you get

Answer 4

i would multiply the second one by \(-4\) to get rid of the \(x\) term

Answer 5

y=1?

Answer 6

yup

Answer 7

x=0?

Answer 8

yes
zero is a perfectly good number

Answer 9

ok so (0,1)

Answer 10

yes

Answer 11

\(\large \begin{array}{llll}
12x+5y=5&&\quad \cancel{12x}+5y=5\\
3x+8y=8&{\color{red}{ \times -4}}\implies &\cancel{-12x}-32y=-32\\
\hline\\
&&\qquad\qquad \square =\qquad\square
\end{array}\)